Problem: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(-6t,4t^{3}\right)$ for time $t\geq 0$. At $t=2$, the particle is at the point $(-2,9)$. What is the particle's position at $t=4$ ? $($
Solution: To find the particle's position at $t=4$, we need to find its horizontal displacement $\Delta x$ and its vertical displacement $\Delta y$, and add those to its initial position $(-2,9)$ : $\text{Position at }t=4\text{: }(-2+\Delta x,9+\Delta y)$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between $t=2$ and $t=4$ : $\Delta x=\int_{2}^{4} -6t\,dt=-36$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between $t=2$ and $t=4$ : $\Delta y=\int_{2}^{4} 4t^{3}\,dt=240$ Now we can find the particle's position: $\begin{aligned} &\phantom{=}(-2+\Delta x,9+\Delta y) \\\\ &=\left(-2+\left(-36\right),9+240\right) \\\\ &=(-38,249) \end{aligned}$ In conclusion, particle's position at $t=4$ is $(-38,249)$.